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Let $F$ be a number field and $E/F$ an elliptic curve with CM by an order $\mathcal{O}$ in a quadratic imaginary field $K$. Let us suppose that $K\subseteq F$. Let $p$ be a prime that splits in $\mathcal{O}$, i.e., $p\mathcal{O}=\wp \overline{\wp}$, and let $\mathfrak{P}$ (resp. $\overline{\mathfrak{P}}$) be a prime of $F$ above $\wp$ (above $\overline{\wp}$), such that $E/F$ has good (ordinary) reduction at $\mathfrak{P}$ (at $\overline{\mathfrak{P}}$). Let $E[p]$ be the group of $p$-torsion points (over $\overline{F}$). Then, $E[p]=E[\wp]\oplus E[\overline{\wp}]$ is split as a $\text{Gal}(\overline{F}/F)$-module.

Hence, under these hypotheses, $E/F$ admits two $F$-rational isogenies of degree $p$, namely $$E \to E/E[\wp], \text{ and } E\to E/E[\overline{\wp}],$$ with kernels $E[\wp]$ and $E[\overline{\wp}]$, respectively, both of order $p$.

On the other hand, the reduction map $\bmod \mathfrak{P}$ induces an exact sequence $$0\to X_\mathfrak{P} \to E(\overline{F})[p] \to E(\mathcal{O}_F/\mathfrak{P})[p]\to 0,$$ and the kernel $X_\mathfrak{P}$ is of order $p$ and it's Galois invariant. Similarly, we have a kernel $X_{\overline{\mathfrak{P}}}$ attached to the similar sequence for reduction $\bmod \overline{\mathfrak{P}}$.

Q: Is $X_\mathfrak{P}=E[\wp]$? Or is $X_\mathfrak{P}=E[\overline{\wp}]$? And why?

I think that $X_\mathfrak{P}=E[\wp]$ because $E[\wp]\cong \wp^{-1}\mathcal{O}/\mathcal{O} \subseteq \mathbb{C}/\mathcal{O}$ and it seems to me that an element of that form should reduce to the origin... but that's not a proof. I feel that the main theorem of CM should be useful here but I am not sure how to use it to conclude what I want. Thanks!

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First, let's show that we may increase the base field $F$ by a finite extension if we need to. Suppose $L/F$ is a finite extension of number fields, base-extend $E$ to be defined over $L$, let $\mathcal{P}$ be a prime of $L$ above your $\mathfrak{P}$ of $F$ (above $\wp$ of $\mathcal{O}$). Since $E/F$ has good (ordinary) reduction at $\mathfrak{P}$, the curve $E/L$ has good (ordinary) reduction at $\mathcal{P}$. Now suppose that the kernel of reduction $\bmod \mathcal{P}$ of $E(\overline{L})[p]$ is $X_\mathcal{P}=E[\wp]$. Since $E$ is originally defined over $F$, we have that $E(\overline{F})[p]\cong E(\overline{L})[p]$, and if a point $R$ reduces to the origin modulo $\mathfrak{P}$, then it also reduces to the origin modulo $\mathcal{P}$, because $\mathcal{P}$ divides $\mathfrak{P}$. Hence, $X_\mathfrak{P}\subseteq X_\mathcal{P}$. Since $|X_\mathfrak{P}|=|X_\mathcal{P}|=p$, we conclude $X_\mathfrak{P}=X_\mathcal{P}=E[\wp]$ as desired.

Ok, now on to the actual proof. Let $E\cong \mathbb{C}/\Lambda$ and let $E'\cong E/E[\wp] \cong \mathbb{C}/\wp^{-1}\Lambda$, so that the isogeny $\phi:\mathbb{C}/\Lambda \to \mathbb{C}/\wp^{-1}\Lambda$ has kernel $E[\wp]\cong \wp^{-1}\Lambda/\Lambda$. Notice that $E[\wp]$ is $\text{Gal}(\overline{F}/F)$-stable, so $E'=E/E[\wp]$ is also defined over $F$. However, the curve $E'$ may not have good reduction at $\mathfrak{P}$, but we can find a finite extension of $F$ such that $E'$ has good reduction at a prime above $\mathfrak{P}$. By our previous lemma this is ok, so without loss of generality let us assume that both $E$ and $E'$ have good (ordinary) reduction modulo $\mathfrak{P}$. Now one can show that the reduction of $\phi \bmod \mathfrak{P}$ is inseparable (this is done in Silverman's "Advanced topics in the arithmetic of elliptic curves", Ch II, $\S 4$, p. 126-127), and therefore the reduction of $\phi$ is essentially a $q$th power Frobenius map (where $q$ is a power of $p$), i.e., $$\tilde{\phi}: \tilde{E} \to \tilde{E}^{(q)} \to \tilde{E}'$$ where the map on the left is $q$th power Frobenius, and the second map is an isomorphism. Moreover, the diagram \begin{align*} & E \to E' \\ & \downarrow \quad \quad \downarrow \\ & \tilde{E} \to \tilde{E'} \end{align*} is commutative. Since $E[\wp]$ is the kernel of the top arrow (which is $\phi$), we must have that $\tilde{\phi}(E[\wp]\bmod \mathfrak{P})=0$. However, the only point mapped to $0$ by the $q$th power Frobenius is $0$ itself, and the isomorphism $\tilde{E}^{(q)}\to \tilde{E'}$ maps $0$ to $0$, so we conclude that $E[\wp] \bmod \mathfrak{P} = 0$, i.e., $X_\mathfrak{P}=E[\wp]$ as desired.

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Thank you for your detailed answer! –  unity Jun 17 '13 at 14:46
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